Find the mean and variance for each of the data : First n natural numbers

Explanation

For first `n` natural numbers
sum `sumn=1+2+3+……….+n=(n(n+1))/2`
`:.` Their mean `bar x=(sumn)/n=((n(n+1))/2)/n=(n+1)/2`
sum of squares of first `barn` natural numbers
`sumn^(2)=1^(2)+2^(2)+3^(2)+………..+n^(2)=(n(n+1)(2n+1))/6`
Now variance `sigma^(2)=(sum n^(2))/n-((sumn)/n)^(2)`
`=(n(n+1)(2n+1))/(6n)-((n+1)/2)^(2)`
`=((n+1)[2(2n+1)-3(n+1)])/12`
`=((n+1)(n-1))/12=(n^(2)-1)/12`